Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, k\neq 0$. $\dfrac{{(r^{-3})^{-3}}}{{(r^{-3}k^{5})^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-3}}$ to the exponent ${-3}$ . Now ${-3 \times -3 = 9}$ , so ${(r^{-3})^{-3} = r^{9}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-3}k^{5})^{3} = (r^{-3})^{3}(k^{5})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-3})^{-3}}}{{(r^{-3}k^{5})^{3}}} = \dfrac{{r^{9}}}{{r^{-9}k^{15}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{9}}}{{r^{-9}k^{15}}} = \dfrac{{r^{9}}}{{r^{-9}}} \cdot \dfrac{{1}}{{k^{15}}} = r^{{9} - {(-9)}} \cdot k^{- {15}} = r^{18}k^{-15}$.